Question

If alpha and beta are the zeroes of the polynomial x^2+x-2, find the value of (1/alpha)-(1/beta)

Answer 1

Answer :-

$\implies \boxed{ \frac{1}{ \alpha} - \frac{1}{ \beta} = - \frac{3}{2} } \:$

To Find :-

$\implies \frac{1}{ \alpha} - \frac{1}{ \beta}$

Explanation :-

We have

→ x² + x - 2 = 0

$\to \boxed{ \sf{sum \: of \: root \: = - \frac{ coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }} \\ \therefore \: \\ \to \boxed{ \alpha + \beta = - 1} \\ \\ \: \to \boxed{\sf{ product \: of \: roots = \frac{ constant}{coefficient \: of \: {x}^{2} } }} \\ \therefore \: \\ \to \boxed{ \alpha \: \beta = - 2}$

Hence ,

$\leadsto \: \frac{1}{ \alpha} - \frac{1}{ \beta} \\ \\ \leadsto \: \frac{ \beta - \alpha}{ \alpha \: \beta} \\ \\ \leadsto \: \frac{ \sqrt{{( \beta + \alpha)}^{2} - 4 \alpha \beta}}{ \alpha \beta} \\ \\ \leadsto \: \frac{ \sqrt{ { ( - 1)}^{2} - 4 \times ( - 2) } }{ - 2} \\ \\ \leadsto \: \frac{ \sqrt{1 + 8} }{ - 2} \\ \\ \leadsto \: \frac{ \sqrt{9} }{ - 2} \\ \\ \leadsto \: \frac{ - 3}{2} \\ \therefore \: \\ \\ \implies \boxed{ \frac{1}{ \alpha} - \frac{1}{ \beta} = - \frac{3}{2} }$